Divide the following complex numbers. $ \dfrac{23-2i}{-3-2i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3+2i}$ $ \dfrac{23-2i}{-3-2i} = \dfrac{23-2i}{-3-2i} \cdot \dfrac{{-3+2i}}{{-3+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(23-2i) \cdot (-3+2i)} {(-3-2i) \cdot (-3+2i)} = \dfrac{(23-2i) \cdot (-3+2i)} {(-3)^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(23-2i) \cdot (-3+2i)} {(-3)^2 - (-2i)^2} = $ $ \dfrac{(23-2i) \cdot (-3+2i)} {9 + 4} = $ $ \dfrac{(23-2i) \cdot (-3+2i)} {13} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({23-2i}) \cdot ({-3+2i})} {13} = $ $ \dfrac{{23} \cdot {(-3)} + {-2} \cdot {(-3) i} + {23} \cdot {2 i} + {-2} \cdot {2 i^2}} {13} $ Evaluate each product of two numbers. $ \dfrac{-69 + 6i + 46i - 4 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{-69 + 6i + 46i + 4} {13} = \dfrac{-65 + 52i} {13} = -5+4i $